Simplify; express your answer in exponential form. Assume $z\neq 0, r\neq 0$. $\dfrac{{(zr^{-1})^{4}}}{{(z^{2}r^{-1})^{-2}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(zr^{-1})^{4} = (z)^{4}(r^{-1})^{4}}$ On the left, we have ${z}$ to the exponent ${4}$ . Now ${1 \times 4 = 4}$ , so ${(z)^{4} = z^{4}}$ Apply the ideas above to simplify the equation. $\dfrac{{(zr^{-1})^{4}}}{{(z^{2}r^{-1})^{-2}}} = \dfrac{{z^{4}r^{-4}}}{{z^{-4}r^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{z^{4}r^{-4}}}{{z^{-4}r^{2}}} = \dfrac{{z^{4}}}{{z^{-4}}} \cdot \dfrac{{r^{-4}}}{{r^{2}}} = z^{{4} - {(-4)}} \cdot r^{{-4} - {2}} = z^{8}r^{-6}$